20105ee115A_1_hw4_solns

20105ee115A_1_hw4_solns - Initially Q = √ 4 = 2 ܥ E = ...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
Q1) In this circuit the switches 1, 2 and 3 has been closed for a long time at t=0s, the switches 1,2 and 3 change their state to open while the switches 4,5 and 6 close. What is the final value of Vout? (you can model switches when they are closed as a resistor R and when they are off as an open circuit) The equivalent circuit before t=0 is At t= 0, the equivalent circuit is : From loop 1 in the figure, 3C = V out (2C) + V 2 C -> 1 From loop 2 in the figure, C – C = -V 2 (C) + V 1 (c) . Therefore V 1 = V 2
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Applying KVL to the outermost loop, V 1 +V 2 = V out . Hence V 2 = V out /2 Substituting the values of V 1 and V 2 in equation 1, 3 = 2V out + V out /2 V out = 6 / 5 V Q2) In this circuit a nonlinear capacitor with initial charge shown in the picture is connected to a linear capacitor which is initially completely discharged. Find the total energy dissipation on the resistor R after a long time.
Background image of page 2
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Background image of page 4
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Initially, Q = √ 4 = 2 ܥ E = ∫ ܸ݀ݍ = ∫ ݍ ଶ ଶ ଴ dq = 8 / 3 E total = 8/3 – ф When t -> ∞, Q = Q 1 + Q 2 2 = √ܸ + (1 x V) V = 1 ܧ = ∫ ݍ ଶ ଵ ଴ dq + 0.5 x C x V 2 E total, ∞ = 1/3 + 5/2 = 5/6 E discharged = E total , 0 – E total, ∞ = 8/3 – 5/6 = 1.83 J 3) Find the amplitudes of the sinusoids at Vo1 and Vo2 when vi is applied The emitter current I E is fixed at 1 mA. So whatever be the change in the base voltage, an ideal current source keeps the current in its path constant. Hence the small signal change in emitter current i e = 0 Therefore the small signal base current i b = i c / ( β + 1) = 0. Hence v Hence v be = 0 V. Here v b = v i and v e = v o2 . Therefore v o2 = v i. Also collector current i c = α x i e = 0 . Hence v o1 = - i c x R = 0V...
View Full Document

{[ snackBarMessage ]}

Page1 / 4

20105ee115A_1_hw4_solns - Initially Q = √ 4 = 2 ܥ E = ...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online